ĐK:\(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\),ta có :
\(M=\frac{2x-\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(M=\frac{2x-x+2\sqrt{x}+2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(M=\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(M=\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(M=\frac{\sqrt{x}+2}{\sqrt{x}-2}\)
b/ \(x=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(x=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(x=1\).Thay vào M ta được :
\(M=\frac{\sqrt{1}+2}{\sqrt{1}-2}=-3\)
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